1. **State the problem:** Solve the system of linear equations: $$\begin{cases} x_1 + x_2 + 2x_3 = -1 \\ 2x_1 - x_2 + 2x_3 = -4 \\ 4x_1 + x_2 + 4x_3 = -2 \end{cases}$$ 2. **Write the system in matrix form:** $$\begin{bmatrix} 1 & 1 & 2 \\ 2 & -1 & 2 \\ 4 & 1 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -1 \\ -4 \\ -2 \end{bmatrix}$$ 3. **Use elimination or substitution to solve:** - From equation 1: $x_1 = -1 - x_2 - 2x_3$ 4. **Substitute $x_1$ into equations 2 and 3:** - Equation 2: $2(-1 - x_2 - 2x_3) - x_2 + 2x_3 = -4$ - Simplify: $$-2 - 2x_2 - 4x_3 - x_2 + 2x_3 = -4$$ $$-2 - 3x_2 - 2x_3 = -4$$ $$-3x_2 - 2x_3 = -2$$ - Equation 3: $4(-1 - x_2 - 2x_3) + x_2 + 4x_3 = -2$ - Simplify: $$-4 - 4x_2 - 8x_3 + x_2 + 4x_3 = -2$$ $$-4 - 3x_2 - 4x_3 = -2$$ $$-3x_2 - 4x_3 = 2$$ 5. **Now solve the system:** $$\begin{cases} -3x_2 - 2x_3 = -2 \\ -3x_2 - 4x_3 = 2 \end{cases}$$ 6. **Subtract the first from the second:** $$(-3x_2 - 4x_3) - (-3x_2 - 2x_3) = 2 - (-2)$$ $$-3x_2 - 4x_3 + 3x_2 + 2x_3 = 4$$ $$-2x_3 = 4$$ $$x_3 = -2$$ 7. **Substitute $x_3 = -2$ into first equation:** $$-3x_2 - 2(-2) = -2$$ $$-3x_2 + 4 = -2$$ $$-3x_2 = -6$$ $$x_2 = 2$$ 8. **Substitute $x_2 = 2$ and $x_3 = -2$ into $x_1 = -1 - x_2 - 2x_3$:** $$x_1 = -1 - 2 - 2(-2) = -1 - 2 + 4 = 1$$ **Final solution:** $$\boxed{x_1 = 1, \quad x_2 = 2, \quad x_3 = -2}$$